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For x in (0,pi) the equation sinx+2sin2x...

For `x in (0,pi)` the equation `sinx+2sin2x-sin3x=3` has

A

infinitely many solutions

B

three solutions

C

one solution

D

no solution

Text Solution

Verified by Experts

The correct Answer is:
D
`sin x +2 sin 2x- sin 3x=3`
`= sin x+4 sin x cos x -3 sin x+4 sin^(3) x=3`
`rArr sin x [-2+4 cos x+4(1-cos^(2) x)]=3`
`rArr sin x [2-(4 cos^(2) x-4 cos x+1)+1]=3`
`rArr 3-(2 cos x-1)^(2) = 3 cosec x`
Now `R.H.S. ge 3`
But `L.H.S. lt 3`
Hence, no solution.
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